The Arithmetic Mean and Median of the given set of data ( 2, 5, 13, 15, 19, 21 ) are 12.5 and 14 respectively.
<h3>What is Arithmetic mean?</h3>
Arithmetic mean is simply the average of a given set numbers. It is determined by dividing the sum of a given set number by their number of appearance.
Mean = Sum total of the number ÷ n
Where n is number of numbers
Median is the middle number in the data set.
Given the sets;
Mean = Sum total of the number ÷ n
Mean = (2 + 5 + 13 + 15 + 19 + 21) ÷ 6
Mean = 75 ÷ 6
Mean = 12.5
Median is the middle number in the data set.
Median = ( 13 + 15 ) ÷ 2
Median = 14
Therefore, the Arithmetic Mean and Median of the given set of data ( 2, 5, 13, 15, 19, 21 ) are 12.5 and 14 respectively.
Learn more about arithmetic mean here: brainly.com/question/13000783
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We can set up an equation to solve this problem. I am setting the number of marbles in a red jar to R.
R + R + R - 16 = 41
We solve this by adding 16 to both sides and combining all of the R terms.. This gives us:
3R = 57
We can finish this problem by dividing both sides by 3.
R = 19. So, there are 19 marbles in a red jar.
We can easily figure out how many marbles are in a blue jar by subtracting the total amount of marbles in 2 red jars from the total amount of marbles. I am setting the amount of marbles in a blue jar to B.
41 - 19*2 = B
B = 3
So, there are 3 marbles in a blue jar and 19 marbles in a red jar.
Answer:
Using trigonometric ratio:
From the given statement:
and sin < 0
⇒
lies in the 3rd quadrant.
then;
Using trigonometry identities:
Substitute the given values we have;
Since, sin < 0
⇒
now, find 
:
Substitute the given values we have;
Therefore, the exact value of:
(a)
(b)
Answer:
James bought 11 good tickets and 5 bad tickets.
Step-by-step explanation:
Given that:
Cost of each good ticket = $8
Cost of each bad ticket = $5
Total amount spent = $113
Total tickets bought = 16
Let,
x be the number of good tickets bought
y be the number of bad tickets bought
x+y=16 Eqn 1
8x+5y=113 Eqn 2
Multiplying Eqn 1 by 5
5(x+y=16)
5x+5y=80 Eqn 3
Subtracting Eqn 3 from Eqn 2
(8x+5y)-(5x+5y)=113-80
8x+5y-5x-5y=33
3x=33
Dividing both sides by 3
Putting x=11 in Eqn 1
11+y=16
y=16-11
y=5
Hence,
James bought 11 good tickets and 5 bad tickets.
